Question

The minimum intensity of audibility of sound is ${10^{ - 12}}watt/{m^2}$. If the intensity of sound is ${10^{ - 9}}watt/{m^2}$,then calculate the intensity level of this sound in decibels.

Answer 1

Hint: The intensity of a sound tells how loud or how much power is carried by the sound per unit area. A common notation to represent the sound intensity is the decibels. It uses the logarithm of 10 and a ratio of reference intensity to the intensity of the given sound.
Formula used
$\beta = 10\log \dfrac{I}{{{I_0}}}$
Where $\beta $is the intensity in decibels.
$I$ is the intensity of sound
${I_0}$ is the minimum or reference intensity.

Complete step by step solution:
 It is given in the question that,
Minimum audible intensity of sound${I_0} = {10^{ - 12}}W/{m^2}$
The given intensity of sound $I = {10^{ - 9}}W/{m^2}$
The intensity level in decibels is given by,
$\beta = 10\log \dfrac{I}{{{I_0}}}$
On putting the values,
$\beta = 10\log \dfrac{{{{10}^{ - 9}}}}{{{{10}^{ - 12}}}}$
$\beta = 10\log {10^3}$
Using the property of logarithm,
$\log {a^x} = x\log a$
We have,
$\beta = 3 \times 10\log 10$
We know that,
${\log _{10}}10 = 1$
Therefore,
$\beta = 30$
The intensity is $30dB$.

Additional information It is known that the loudness of sound depends on its intensity. But the sensitivity of the sound in human ears does not vary linearly, it does so by a factor of 10 this is why the decibel level was started to be used. If you want to double the volume of a sound, you will have to increase its intensity by 10 times. Thus the decibel level fits the representation of sound in terms of human use.

Note: It is to be kept in mind that the logarithm used in the calculations of Decibels has the base as $10$, and not in base $e$. So if the intensity ratios are given in powers of 10 the decibel simply becomes the value of the exponent. The $\beta $ is defined as a ratio between two intensities, which is why it does not have a unit in terms of dimensions. But the reference and the observed Intensities have the units of $W/{m^2}$and dimension of mass and time.